BE Civil Engineering (IOE, TU) Hydraulics (IOE, CE 604) Question Paper 2077 Nepal

Given: $b = 6.0\ \text{m}$, side slope $z = 2$ (H:V), $S_0 = 0.0009$, $n = 0.018$, $y = 1.5\ \text{m}$, $g = 9.81\ \text{m/s}^2$.

(a) Discharge (Manning's equation)

Flow area:

$$A = (b + zy)y = (6.0 + 2\times1.5)\times1.5 = (6.0+3.0)\times1.5 = 13.5\ \text{m}^2$$

Wetted perimeter:

$$P = b + 2y\sqrt{1+z^2} = 6.0 + 2\times1.5\sqrt{1+4} = 6.0 + 3.0\times2.2360680 = 6.0 + 6.7082039 = 12.7082\ \text{m}$$

Hydraulic radius:

$$R = \frac{A}{P} = \frac{13.5}{12.7082} = 1.06231\ \text{m}$$

Manning's equation:

$$Q = \frac{1}{n}A R^{2/3} S_0^{1/2}$$

$R^{2/3} = 1.06231^{2/3}$. Since $\ln 1.06231 = 0.060456$, $\tfrac{2}{3}\times0.060456 = 0.040304$, $e^{0.040304} = 1.041128$.

$S_0^{1/2} = \sqrt{0.0009} = 0.03$.

$$Q = \frac{1}{0.018}\times13.5\times1.041128\times0.03 = 55.5556\times13.5\times1.041128\times0.03$$ $$Q = 55.5556\times13.5 = 750.000;\quad 750.000\times1.041128 = 780.846;\quad \times0.03$$ $$\boxed{Q = 23.43\ \text{m}^3/\text{s}}$$

(b) Boundary shear stress and flow state

Average boundary shear stress:

$$\tau_0 = \gamma R S_0 = 9810\times1.06231\times0.0009 = 9.379\ \text{N/m}^2$$ $$\boxed{\tau_0 \approx 9.38\ \text{Pa}}$$

Flow state — mean velocity:

$$V = \frac{Q}{A} = \frac{23.43}{13.5} = 1.7356\ \text{m/s}$$

Top width $T = b + 2zy = 6.0 + 2\times2\times1.5 = 12.0\ \text{m}$; hydraulic (mean) depth $D = A/T = 13.5/12.0 = 1.125\ \text{m}$.

Froude number:

$$Fr = \frac{V}{\sqrt{gD}} = \frac{1.7356}{\sqrt{9.81\times1.125}} = \frac{1.7356}{\sqrt{11.0363}} = \frac{1.7356}{3.32209} = 0.5224$$

Since $Fr = 0.52 < 1$, the flow is subcritical.

(c) Conveyance

Writing $Q = K S_0^{1/2}$, the conveyance is $K = \dfrac{1}{n}A R^{2/3}$. It is a purely geometric–roughness measure of a section's ability to carry flow: it groups all terms that depend only on the channel shape, size and roughness, separating them from the energy slope. For a given $K$, discharge varies with $\sqrt{S_0}$. Here $K = Q/S_0^{1/2} = 23.43/0.03 = 781.0\ \text{m}^3/\text{s}$.

Given: $b = 4.0\ \text{m}$, $Q = 12\ \text{m}^3/\text{s}$, $g = 9.81\ \text{m/s}^2$. Unit discharge $q = Q/b = 12/4.0 = 3.0\ \text{m}^2/\text{s}$.

(a) Critical-flow condition

Specific energy: $E = y + \dfrac{V^2}{2g} = y + \dfrac{q^2}{2gy^2}$ (per unit width, rectangular).

For minimum $E$ at fixed $q$, set $\dfrac{dE}{dy}=0$:

$$\frac{dE}{dy} = 1 - \frac{q^2}{g y^3} = 0 \;\Rightarrow\; y_c = \left(\frac{q^2}{g}\right)^{1/3}$$

At this depth $\dfrac{q^2}{g y_c^3} = 1$, i.e. $\dfrac{V_c^2}{g y_c} = 1 \Rightarrow Fr=1$ (critical). Then

$$E_c = y_c + \frac{q^2}{2g y_c^2} = y_c + \frac{y_c}{2} = \tfrac{3}{2}y_c.$$

(b) Critical depth, velocity, minimum specific energy

$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.0^2}{9.81}\right)^{1/3} = \left(\frac{9.0}{9.81}\right)^{1/3} = (0.917431)^{1/3}$$

$\ln 0.917431 = -0.086178$; $/3 = -0.028726$; $e^{-0.028726}=0.971682$.

$$\boxed{y_c = 0.9717\ \text{m}}$$ $$V_c = \frac{q}{y_c} = \frac{3.0}{0.9717} = 3.0874\ \text{m/s}\quad\Rightarrow\boxed{V_c \approx 3.09\ \text{m/s}}$$ $$E_c = \tfrac{3}{2}y_c = 1.5\times0.9717 = \boxed{1.4575\ \text{m}}$$

(c) Specific energy at $y = 1.8\ \text{m}$ and alternate depth

$$V = \frac{q}{y} = \frac{3.0}{1.8} = 1.6667\ \text{m/s},\qquad \frac{V^2}{2g} = \frac{1.6667^2}{2\times9.81} = \frac{2.77778}{19.62} = 0.14157\ \text{m}$$ $$E = 1.8 + 0.14157 = \boxed{1.9416\ \text{m}}\quad(> E_c,\ \text{so depth is valid; flow is subcritical}).$$

Alternate depth $y'$ satisfies $E = y' + \dfrac{q^2}{2g y'^2}$, i.e.

$$1.9416 = y' + \frac{9.0}{19.62\,y'^2} = y' + \frac{0.458716}{y'^2}.$$

Solve iteratively for the supercritical root ($y' < y_c$):

• $y'=0.6$: $0.6 + 0.458716/0.36 = 0.6+1.27421=1.87421$ (low)
• $y'=0.58$: $0.58 + 0.458716/0.3364 = 0.58+1.36361=1.94361$ (≈)
• $y'=0.581$: $0.581 + 0.458716/0.337561 = 0.581+1.358918=1.939918$

Interpolating, $y' \approx 0.5804\ \text{m}$. Check: $0.5804 + 0.458716/0.336864 = 0.5804+1.361729 = 1.94213$.

$$\boxed{y' \approx 0.580\ \text{m}\ (\text{supercritical alternate depth})}$$

Given: $b = 5.0\ \text{m}$, $Q = 30\ \text{m}^3/\text{s}$, $y_1 = 0.6\ \text{m}$, $g = 9.81\ \text{m/s}^2$.

Unit discharge $q = 30/5.0 = 6.0\ \text{m}^2/\text{s}$.

(a) Upstream Froude number

$$V_1 = \frac{q}{y_1} = \frac{6.0}{0.6} = 10.0\ \text{m/s}$$ $$Fr_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{10.0}{\sqrt{9.81\times0.6}} = \frac{10.0}{\sqrt{5.886}} = \frac{10.0}{2.42611} = 4.1218$$ $$\boxed{Fr_1 = 4.12}$$

Since $4.5 > Fr_1 > 2.5$, this is a steady (oscillating-to-steady boundary) jump — strictly $2.5<Fr_1<4.5$ is the oscillating range; with $Fr_1=4.12$ it is a well-developed oscillating jump approaching the steady type.

(b) Sequent depth (Belanger equation)

$$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)$$ $$8Fr_1^2 = 8\times4.1218^2 = 8\times16.9893 = 135.914;\quad 1+135.914 = 136.914$$ $$\sqrt{136.914} = 11.7011$$ $$\frac{y_2}{y_1} = \frac{1}{2}(11.7011-1) = \frac{10.7011}{2} = 5.35057$$ $$y_2 = 5.35057\times0.6 = \boxed{3.210\ \text{m}}$$

(c) Energy loss and efficiency

Energy loss in a rectangular jump:

$$\Delta E = \frac{(y_2-y_1)^3}{4 y_1 y_2} = \frac{(3.210-0.6)^3}{4\times0.6\times3.210} = \frac{(2.610)^3}{7.704} = \frac{17.7796}{7.704} = 2.3079\ \text{m}$$ $$\boxed{\Delta E \approx 2.31\ \text{m}}$$

Specific energies:

$$E_1 = y_1 + \frac{V_1^2}{2g} = 0.6 + \frac{10.0^2}{19.62} = 0.6 + 5.0968 = 5.6968\ \text{m}$$ $$V_2 = \frac{q}{y_2} = \frac{6.0}{3.210} = 1.8692\ \text{m/s};\quad E_2 = 3.210 + \frac{1.8692^2}{19.62} = 3.210 + \frac{3.49391}{19.62} = 3.210 + 0.17808 = 3.3881\ \text{m}$$

Check: $E_1 - E_2 = 5.6968 - 3.3881 = 2.3087\ \text{m} \approx \Delta E$ (consistent).

Efficiency:

$$\frac{E_2}{E_1} = \frac{3.3881}{5.6968} = 0.5947 \Rightarrow \boxed{59.5\%}$$

The jump dissipates about $40.5\%$ of the upstream specific energy.

(a) Derivation of the GVF equation

Total head above a datum: $H = z + y + \dfrac{V^2}{2g}$.

Assumptions: (i) steady flow; (ii) gradual depth change so pressure is hydrostatic and streamlines nearly parallel; (iii) the head loss at a section equals that of uniform flow at the same depth/velocity, so $S_f$ from Manning/Chezy applies; (iv) channel slope small ($\cos\theta\approx1$); (v) prismatic channel, $\alpha=1$; (vi) constant $n$.

Differentiate w.r.t. $x$ (taking $x$ positive downstream):

$$\frac{dH}{dx} = \frac{dz}{dx} + \frac{dy}{dx} + \frac{d}{dx}\!\left(\frac{V^2}{2g}\right)$$

Now $\dfrac{dH}{dx} = -S_f$ (energy slope), $\dfrac{dz}{dx} = -S_0$ (bed slope). Also $\dfrac{d}{dx}\!\left(\dfrac{V^2}{2g}\right) = -Fr^2\dfrac{dy}{dx}$ for a given $Q$. Substituting:

$$-S_f = -S_0 + \frac{dy}{dx}(1 - Fr^2)\;\Rightarrow\; \boxed{\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2}}$$

(b) Normal and critical depths; slope classification

Wide channel ⇒ $R \approx y$. Manning (per unit width): $q = \dfrac{1}{n} y^{5/3} S_0^{1/2}$.

$$y_n = \left(\frac{q\,n}{S_0^{1/2}}\right)^{3/5} = \left(\frac{4\times0.020}{\sqrt{0.004}}\right)^{3/5} = \left(\frac{0.08}{0.0632456}\right)^{3/5} = (1.264911)^{3/5}$$

$\ln 1.264911 = 0.234923$; $\times0.6 = 0.140954$; $e^{0.140954} = 1.151367$.

$$\boxed{y_n = 1.151\ \text{m}}$$

Critical depth (wide rectangular): $y_c = (q^2/g)^{1/3} = (16/9.81)^{1/3} = (1.630989)^{1/3}$. $\ln 1.630989 = 0.489149$; $/3 = 0.163050$; $e^{0.163050} = 1.177099$.

$$\boxed{y_c = 1.177\ \text{m}}$$

Since $y_n (1.151) < y_c (1.177)$, the normal flow is supercritical ⇒ the slope is steep (S).

Possible profiles on a steep slope: S1 (for $y > y_c$), S2 (for $y_c > y > y_n$), S3 (for $y < y_n$).

ASCII sketch (steep slope zones):

         ~~~~~~ S1  (y > y_c)
----------------------------- critical depth line  y_c=1.177
   \___ S2 (y_n<y<y_c)
----------------------------- normal depth line    y_n=1.151
   ___/  S3 (y < y_n)
============================= bed (steep)

(c) Behaviour at $y = 1.2\ \text{m}$

Here $y = 1.2\ \text{m} > y_c = 1.177 > y_n = 1.151$, so this is the S1 zone. Numerator $S_0 - S_f$: since $y > y_n$, the actual depth exceeds normal ⇒ $S_f < S_0$ ⇒ numerator $>0$. Denominator $1 - Fr^2$: since $y > y_c$ the flow is subcritical ⇒ $Fr<1$ ⇒ denominator $>0$. Thus $\dfrac{dy}{dx} > 0$: the water surface rises in the flow direction (S1 backwater profile, asymptotic to the horizontal far upstream-of-an-obstruction and tending away from $y_c$).

Given: $Q = 20\ \text{m}^3/\text{s}$, side slope $z = 1.5$ (H:V), $n = 0.022$, $S_0 = 0.0006$, $g=9.81$.

(a) Most-efficient trapezoidal section

For fixed $A$ and $z$, the section is most efficient when $P$ is minimum (maximises $R$, hence $Q$). $A = (b+zy)y$, $P = b + 2y\sqrt{1+z^2}$. From $A$: $b = A/y - zy$. So

$$P = \frac{A}{y} - zy + 2y\sqrt{1+z^2}.$$

Set $dP/dy = 0$ at constant $A$:

$$\frac{dP}{dy} = -\frac{A}{y^2} - z + 2\sqrt{1+z^2} = 0 \Rightarrow A = y^2\left(2\sqrt{1+z^2} - z\right).$$

Equating to $A=(b+zy)y$ gives the best-section relation:

$$\boxed{b + zy \,=\, y\left(2\sqrt{1+z^2}-z\right)\;\Rightarrow\; b = y\left(2\sqrt{1+z^2}-2z\right).}$$

A consequence is $R = y/2$ (hydraulic radius equals half the depth) for the most efficient trapezoid.

(b) Design

With $z = 1.5$: $\sqrt{1+z^2} = \sqrt{1+2.25} = \sqrt{3.25} = 1.802776$.

Bed width: $b = y(2\times1.802776 - 2\times1.5) = y(3.605551 - 3.0) = 0.605551\,y$.

Area: $A = y^2(2\sqrt{1+z^2} - z) = y^2(3.605551 - 1.5) = 2.105551\,y^2$.

Hydraulic radius $R = y/2 = 0.5y$.

Manning: $Q = \dfrac{1}{n} A R^{2/3} S_0^{1/2}$.

$$20 = \frac{1}{0.022}\,(2.105551\,y^2)(0.5y)^{2/3}(0.0006)^{1/2}.$$

$(0.5)^{2/3}$: $\ln0.5=-0.693147$, $\times\tfrac23=-0.462098$, $e^{-0.462098}=0.629961$. So $(0.5y)^{2/3}=0.629961\,y^{2/3}$. $S_0^{1/2}=\sqrt{0.0006}=0.0244949$.

$$20 = \frac{1}{0.022}\times2.105551\times0.629961\times0.0244949\;y^{2+2/3}$$

Coefficient: $\frac{1}{0.022}=45.4545$; $45.4545\times2.105551 = 95.7069$; $\times0.629961 = 60.2898$; $\times0.0244949 = 1.476788$.

$$20 = 1.476788\,y^{8/3} \Rightarrow y^{8/3} = \frac{20}{1.476788} = 13.54295.$$

$y = 13.54295^{3/8}$: $\ln13.54295 = 2.605910$; $\times0.375 = 0.977216$; $e^{0.977216} = 2.656832$.

$$\boxed{y = 2.657\ \text{m}}$$

Bed width:

$$b = 0.605551\times2.657 = \boxed{1.609\ \text{m}}$$

Velocity check:

$$A = 2.105551\times2.657^2 = 2.105551\times7.059649 = 14.86489\ \text{m}^2$$ $$V = \frac{Q}{A} = \frac{20}{14.86489} = 1.3455\ \text{m/s}$$

Since $V = 1.35\ \text{m/s} > 1.0\ \text{m/s}$ (permissible non-scouring velocity), the design velocity exceeds the permissible value. The section would scour; the bed slope must be flattened (reduce $S_0$) or the channel lined / widened from the strict best section. (If a higher permissible velocity were allowed, e.g. for a lined channel, the section is acceptable.)

Given: $V_0 = 2.5\ \text{m/s}$, $L = 1200\ \text{m}$, $a = 1200\ \text{m/s}$, $\rho = 1000\ \text{kg/m}^3$, $g = 9.81\ \text{m/s}^2$.

Pipe period: $T = \dfrac{2L}{a} = \dfrac{2\times1200}{1200} = 2.0\ \text{s}$.

(a) Sudden vs gradual closure

Closure is sudden (rapid) if the closure time $t_c \le T = 2L/a$ — the reflected relief wave has not yet returned to the valve, so the full Joukowsky head develops. It is gradual (slow) if $t_c > 2L/a$ — the returning negative wave partially relieves the pressure, so the peak head is smaller than the Joukowsky value.

(b) Joukowsky (instantaneous) pressure rise

$$\Delta p = \rho\,a\,V_0 = 1000\times1200\times2.5 = 3.0\times10^{6}\ \text{Pa} = \boxed{3.0\ \text{MPa}}$$

As head:

$$\Delta H = \frac{a V_0}{g} = \frac{1200\times2.5}{9.81} = \frac{3000}{9.81} = 305.81\ \text{m}\;\Rightarrow\boxed{\Delta H \approx 305.8\ \text{m}}$$

(c) Closure in $t_c = 3.0\ \text{s}$

Since $t_c = 3.0\ \text{s} > T = 2.0\ \text{s}$, the closure is slow (gradual).

Using the (Michaud/Allievi) approximation for slow uniform closure:

$$\Delta H = \frac{2 L V_0}{g\,t_c} = \frac{2\times1200\times2.5}{9.81\times3.0} = \frac{6000}{29.43} = 203.87\ \text{m}$$ $$\boxed{\Delta H \approx 203.9\ \text{m}}$$

This is well below the Joukowsky value, confirming the benefit of slower closure. (Equivalently $\Delta H = \Delta H_{max}\cdot T/t_c = 305.8\times2.0/3.0 = 203.9\ \text{m}$.)

Given (wide channel, $R\approx y$): $q = 3.0\ \text{m}^2/\text{s}$, $S_0 = 0.0005$, $n = 0.025$, $g=9.81$.

Direct step: $\Delta x = \dfrac{E_2 - E_1}{S_0 - \bar S_f}$, where $E = y + \dfrac{q^2}{2g y^2}$ and $\bar S_f$ is the mean friction slope.

Specific energy at each section ($\dfrac{q^2}{2g} = \dfrac{9.0}{19.62} = 0.458716$):

$E_2 - E_1 = 2.294776 - 2.114679 = 0.180097\ \text{m}$.

Friction slope $S_f = \dfrac{n^2 V^2}{R^{4/3}} = \dfrac{n^2 q^2}{y^{10/3}}$ (wide channel, $R=y$).

$n^2 q^2 = 0.025^2\times3.0^2 = 0.000625\times9.0 = 0.005625$.

Section 1: $y^{10/3} = 2.0^{10/3}$. $\ln2=0.693147$, $\times10/3=2.310490$, $e^{2.310490}=10.0794$. $S_{f1} = 0.005625/10.0794 = 5.5807\times10^{-4}$.

Section 2: $y^{10/3} = 2.2^{10/3}$. $\ln2.2=0.788457$, $\times10/3=2.628190$, $e^{2.628190}=13.8497$. $S_{f2} = 0.005625/13.8497 = 4.0614\times10^{-4}$.

Mean: $\bar S_f = \tfrac12(5.5807+4.0614)\times10^{-4} = 4.82105\times10^{-4}$.

Step length:

$$\Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f} = \frac{0.180097}{0.0005 - 0.000482105} = \frac{0.180097}{0.0000178950}$$ $$\Delta x = 10063.8\ \text{m}\;\Rightarrow\boxed{\Delta x \approx 1.01\times10^{4}\ \text{m}\ (\approx 10.06\ \text{km})}$$

The positive $\Delta x$ (with depth rising and $S_f<S_0$, i.e. $y>y_n$) indicates an M1 backwater computed in the upstream direction; the large length reflects the gentle slope and the depth being only slightly above normal.

Given: $L = 3.0\ \text{m}$, $H = 0.45\ \text{m}$, $C_d = 0.62$, $g = 9.81\ \text{m/s}^2$.

(a) Discharge (no approach velocity)

$$Q = \frac{2}{3} C_d L \sqrt{2g}\, H^{3/2}$$

$\sqrt{2g} = \sqrt{19.62} = 4.429446$. $H^{3/2} = 0.45^{1.5}$: $\sqrt{0.45}=0.670820$, $\times0.45 = 0.301869$.

$$Q = \tfrac{2}{3}\times0.62\times3.0\times4.429446\times0.301869$$

$\tfrac23\times0.62 = 0.413333$; $\times3.0 = 1.24$; $\times4.429446 = 5.492513$; $\times0.301869 = 1.65805$.

$$\boxed{Q \approx 1.658\ \text{m}^3/\text{s}}$$

(b) With velocity of approach (one iteration)

Approach area $A = 3.0\times1.2 = 3.6\ \text{m}^2$. Approach velocity $V_a = Q/A = 1.65805/3.6 = 0.46057\ \text{m/s}$. Approach head $h_a = \dfrac{V_a^2}{2g} = \dfrac{0.46057^2}{19.62} = \dfrac{0.212125}{19.62} = 0.010812\ \text{m}$.

Corrected discharge:

$$Q = \frac{2}{3}C_d L\sqrt{2g}\left[(H+h_a)^{3/2} - h_a^{3/2}\right]$$

$(H+h_a) = 0.460812$; $(0.460812)^{3/2}$: $\sqrt{0.460812}=0.678832$, $\times0.460812 = 0.312819$. $h_a^{3/2} = (0.010812)^{1.5}$: $\sqrt{0.010812}=0.103981$, $\times0.010812 = 0.0011242$. Bracket $= 0.312819 - 0.0011242 = 0.311695$.

$$Q = 5.492513\times0.311695 = 1.71200\ \text{m}^3/\text{s}$$ $$\boxed{Q \approx 1.712\ \text{m}^3/\text{s}}$$

(about $3.3\%$ higher than the no-approach value).

(c) Two uses of weirs

1. Flow measurement — a calibrated weir gives discharge from a single head reading $H$.
2. Water-level / flow regulation — weirs raise and maintain upstream water levels for diversion, head regulation at canal intakes, and act as overflow/escape structures (e.g., spillways, drop structures) to pass surplus flow safely.

Given: $b = 4.0\ \text{m}$, $y_1 = 1.0\ \text{m}$ (initial), $V_1 = 1.5\ \text{m/s}$, $y_2 = 1.6\ \text{m}$ (after surge), $g = 9.81\ \text{m/s}^2$. Positive surge moving upstream with absolute speed $V_w$ (take upstream as the surge direction; flow is downstream).

Let the surge move upstream with absolute speed $C$ (magnitude). Use control-volume relations on the moving surge. For a surge advancing upstream against a flow, with $V_w = C$ (speed of surge), the relations are:

Continuity (relative to surge):

$$(V_1 + V_w)y_1 = (V_2 + V_w)y_2$$

Momentum / surge celerity (positive surge): relative approach velocity equals

$$(V_1 + V_w) = \sqrt{\frac{g\,y_2}{2 y_1}(y_1 + y_2)}$$

(a) Surge speed

$$V_1 + V_w = \sqrt{\frac{9.81\times1.6}{2\times1.0}(1.0+1.6)} = \sqrt{\frac{15.696}{2.0}\times2.6} = \sqrt{7.848\times2.6} = \sqrt{20.4048} = 4.51717\ \text{m/s}$$ $$V_w = 4.51717 - V_1 = 4.51717 - 1.5 = 3.0172\ \text{m/s}$$ $$\boxed{V_w \approx 3.02\ \text{m/s (upstream)}}$$

(b) Velocity behind the surge From continuity:

$$V_2 + V_w = \frac{(V_1+V_w)\,y_1}{y_2} = \frac{4.51717\times1.0}{1.6} = 2.823231\ \text{m/s}$$ $$V_2 = 2.823231 - V_w = 2.823231 - 3.0172 = -0.1940\ \text{m/s}$$ $$\boxed{V_2 \approx -0.19\ \text{m/s}}$$

The small negative value means the water behind the surge is nearly stagnant / very slightly reversed — consistent with a strong downstream gate restriction.

Check (continuity, absolute): discharge change is absorbed by the rising level; relative discharge $(V_1+V_w)y_1 = 4.51717\ \text{m}^2/\text{s}$ and $(V_2+V_w)y_2 = 2.823231\times1.6 = 4.51717\ \text{m}^2/\text{s}$ ✓.

(c) Moving hydraulic jump?

Yes. A positive surge is essentially a hydraulic jump translating along the channel. In a reference frame moving with the surge at speed $V_w$, the flow is steady and satisfies the same momentum (Belanger-type) relation as a stationary jump, with the abrupt rise from $y_1$ to $y_2$. Here the relative upstream Froude number $Fr_r = (V_1+V_w)/\sqrt{g y_1} = 4.51717/\sqrt{9.81} = 4.51717/3.13209 = 1.442 > 1$, confirming supercritical relative inflow — the hallmark of a (moving) hydraulic jump.

Given: $Q = 8\ \text{m}^3/\text{s}$, $b = 2.5\ \text{m}$, $y_1 = 1.4\ \text{m}$, $\Delta z = 0.15\ \text{m}$, $g=9.81$. Unit discharge $q = 8/2.5 = 3.2\ \text{m}^2/\text{s}$.

Upstream specific energy:

$$V_1 = q/y_1 = 3.2/1.4 = 2.285714\ \text{m/s};\quad \frac{V_1^2}{2g} = \frac{2.285714^2}{19.62} = \frac{5.224490}{19.62} = 0.266284\ \text{m}$$ $$E_1 = 1.4 + 0.266284 = 1.666284\ \text{m}$$

Flow is subcritical ($y_1 > y_c$, see below).

Critical depth: $y_c = (q^2/g)^{1/3} = (3.2^2/9.81)^{1/3} = (10.24/9.81)^{1/3} = (1.043833)^{1/3}$. $\ln1.043833 = 0.042896$; $/3 = 0.014299$; $e^{0.014299}=1.014402$. So $y_c = 1.0144\ \text{m}$; $E_c = 1.5 y_c = 1.52160\ \text{m}$.

(a) Depth over hump

Energy over the hump: $E_2 = E_1 - \Delta z = 1.666284 - 0.15 = 1.516284\ \text{m}$.

Note $E_2 = 1.51628\ \text{m} > E_{c} = 1.52160$? Compare: $1.51628 < 1.52160$. This is marginally below $E_c$, which would mean choking. Re-check $E_c$: $E_c = 1.5\times1.014402 = 1.521603$. Indeed $E_2 (1.51628) < E_c (1.52160)$ by $0.0053\ \text{m}$ — the proposed hump just barely chokes the flow.

Because $E_2$ falls a hair below the minimum energy, strictly the flow chokes and $y_1$ must rise. For the value to be physically passable we take the limiting case: the depth over the crest is critical, $y_2 = y_c = 1.014\ \text{m}$. (If losses/idealisation place $E_2$ exactly at $E_c$, the surface passes through critical on the hump.)

$$\boxed{y_2 \approx y_c = 1.014\ \text{m (flow at/near critical on the hump)}}$$

(b) Maximum (choking) hump height

Choking occurs when $E_2 = E_c$, i.e.

$$\Delta z_{max} = E_1 - E_c = 1.666284 - 1.521603 = 0.144681\ \text{m}$$ $$\boxed{\Delta z_{max} \approx 0.145\ \text{m}}$$

Since the proposed $\Delta z = 0.15\ \text{m} > 0.145\ \text{m}$, the hump exceeds the choking height: the upstream depth will increase (back up) and the flow over the crest is critical. This confirms part (a).

(a) Relation between $C$ and $n$

Chezy: $V = C\sqrt{R S_f}$. Manning: $V = \dfrac{1}{n} R^{2/3} S_f^{1/2}$. Equating the two velocity expressions:

$$C\sqrt{R S_f} = \frac{1}{n} R^{2/3} S_f^{1/2} \;\Rightarrow\; C R^{1/2} = \frac{1}{n} R^{2/3}$$ $$\boxed{C = \frac{1}{n} R^{1/6}}$$

Thus Chezy's $C$ depends weakly (one-sixth power) on hydraulic radius, whereas $n$ is treated as constant for a given surface.

(b) Numerical value of $C$

Rectangular: $A = b y = 3.0\times1.0 = 3.0\ \text{m}^2$; $P = b + 2y = 3.0 + 2.0 = 5.0\ \text{m}$; $R = A/P = 3.0/5.0 = 0.6\ \text{m}$. $R^{1/6} = 0.6^{1/6}$: $\ln0.6 = -0.510826$; $/6 = -0.085138$; $e^{-0.085138} = 0.918386$.

$$C = \frac{1}{0.015}\times0.918386 = 66.6667\times0.918386 = 61.226\ \text{m}^{1/2}/\text{s}$$ $$\boxed{C \approx 61.2\ \text{m}^{1/2}/\text{s}}$$

(c) Location of maximum velocity

In an open channel the free surface is subject to secondary currents and air–water interaction (surface resistance, wind/surface tension and the downward-directed secondary flow cells). These secondary (cross-circulation) currents carry low-momentum fluid from the corners and sides toward the centre of the surface and push high-momentum fluid downward, so the point of maximum velocity is depressed to roughly $0.05$–$0.25$ of the depth below the surface (commonly about $0.2 y$ below it for wide channels). Consequently the mean velocity occurs near $0.6$ of the depth from the surface, and the average of velocities at $0.2 y$ and $0.8 y$ approximates the mean — the basis of the standard velocity-measurement rules.